Let the points P(x_(n−1) ,y_(n−1) ), Q(x_n ,y_n ) and R(x_(n+1) ,y_(n+1) ) lies on the curve y = f(x). Prove that y_(n+1) ≈ y_(n−1) + 2h ((dy/dx))


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Question Number 122187 by physicstutes last updated on 14/Nov/20

$Let the points P (x_{n - 1}, y_{n - 1}), Q (x_{n}, y_{n}) and R (x_{n + 1}, y_{n + 1}) lies$ $on the curve y = f (x) . Prove that$ $y_{n + 1} \approx y_{n - 1} + 2 h {(\frac{d y}{d x})}_{n} .$
	Answered by Olaf last updated on 14/Nov/20
	${ ((((dy/dx))_n ≈ ((y_n −y_(n−1) )/(x_n −x_(n−1) )) (1))),((((dy/dx))_n ≈ ((y_(n+1) −y_n )/(x_(n+1) −x_n ))(2))) :} { ((((dy/dx))_n ≈ ((y_n −y_(n−1) )/h) (1))),((((dy/dx))_n ≈ ((y_(n+1) −y_n )/h) (2))) :} (1)+(2) : 2((dy/dx))_n ≈ ((y_(n+1) −y_(n−1) )/h) ⇒ y_(n+1) ≈ y_(n−1) +2h((dy/dx))_n$
	${\begin{cases} {(\frac{d y}{d x})}_{n} \approx \frac{y_{n} - y_{n - 1}}{x_{n} - x_{n - 1}} (1) \\ {(\frac{d y}{d x})}_{n} \approx \frac{y_{n + 1} - y_{n}}{x_{n + 1} - x_{n}} (2) \end{cases}$ ${\begin{cases} {(\frac{d y}{d x})}_{n} \approx \frac{y_{n} - y_{n - 1}}{h} (1) \\ {(\frac{d y}{d x})}_{n} \approx \frac{y_{n + 1} - y_{n}}{h} (2) \end{cases}$ $(1) + (2) :$ $2 {(\frac{d y}{d x})}_{n} \approx \frac{y_{n + 1} - y_{n - 1}}{h}$ $\Rightarrow y_{n + 1} \approx y_{n - 1} + 2 h {(\frac{d y}{d x})}_{n}$
	Commented by physicstutes last updated on 14/Nov/20

	$Thank you sir . . .$ $What if i do this :$ $Assuming P \to 0, R \to 0 such that we can approximate$ $f (x) to a straight line .$ $then; {(\frac{d y}{d x})}_{n} \approx \frac{y_{n + 1} - y_{n - 1}}{x_{n + 1} - x_{n - 1}}$ $where x_{n + 1} - x_{n - 1} = x_{n + 1} - x_{n} + x_{n} - x_{n - 1} = 2 h$ $\Rightarrow {(\frac{d y}{d x})}_{n} \approx \frac{y_{n + 1} - y_{n - 1}}{2 h}$ $\Rightarrow y_{n + 1} \approx y_{n - 1} + 2 h {(\frac{d y}{d x})}_{n}$ $Is my approxation (P \to 0, R \to 0) worth it ?$
	Commented by Olaf last updated on 14/Nov/20

	$Yes sir but when you write$ ${(\frac{d y}{d x})}_{n} = \frac{y_{n + 1} - y_{n - 1}}{x_{n + 1} - x_{n}}, the ratio is not$ $at point x_{n .} In fact you use the result$ $to demonstrate the result .$
	Commented by physicstutes last updated on 14/Nov/20

	$Actually i write,$ ${(\frac{d y}{d x})}_{n} \approx \frac{y_{n + 1} - y_{n - 1}}{x_{n + 1} - x_{n - 1}}$ $considering the points P (x_{n - 1}, y_{n - 1}) and R (x_{n + 1}, y_{n + 1})$ $Now i used the idea that, the step length (or x - differences) is h$ $so the idea is x_{n + 1} - x_{n} = x_{n} - x_{n - 1} = h$ $\Rightarrow x_{n + 1} - x_{n - 1} = x_{n + 1} - x_{n} + x_{n} - x_{n - 1} = 2 h$ $This happens when we solve differential equations numerical .$ $so does this makes more sense ?$
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